1991 AIME Problems/Problem 4

Problem

How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$ ?

Solution

The range of the sine function is $-1 \le y \le 1$ . It is periodic (in this problem) with a period of $\frac{2}{5}$ .

Thus, $-1 \le \frac{1}{5} \log_2 x \le 1$ , and $-5 \le \log_2 x \le 5$ . The solutions for $x$ occur in the domain of $\frac{1}{32} \le x \le 32$ . When $x > 1$ the logarithm function returns a positive value; up to $x = 32$ it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of $x$ ) of the sine curve and another curve that is $< 1$ , so there are $\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154$ values (the subtraction of 6 since all the “intersections” when $x < 1$ must be disregarded). When $y = 0$ , there is exactly $1$ touching point between the two functions: $\left(\frac{1}{5},0\right)$ . When $y < 0$ or $x < 1$ , we can count $4$ more solutions. The solution is $154 + 1 + 4 = \boxed{159}$ .

Solution 2

Notice that the equation is satisfied twice for every sine period (which is $\frac{2}{5}$ ), except in the sole case when the two equations equate to $0$ . In that case, the equation is satisfied twice but only at the one instance when $y=0$ . Hence, it is double-counted in our final solution, so we have to subtract it out. We then compute: $32 \cdot \frac{5}{2} \cdot 2 - 1 = \boxed {159}$

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1991 AIME Problems/Problem 4

Contents

Problem

Solution

Solution 2

See also